First of all, I suck at math. I can figure out tips at a restaurant, and how long it takes to get from point A to point B; other than that, math and I don’t get along. Honestly, one of the reasons I majored in poly sci was because I only had to take two math classes, which were statistics. Which brings me to my question.
What are the chances of getting a 2 on a 6 sided die?
What are the chances of pulling a Jack of Spades from a deck of cards?
My answer: 50/50. Either you will or you won’t.
Now, I know the correct answer to the questions are 1 in 6 and 1 in 52. I’m always told I’m wrong with the 50/50 answer. Why? If I roll a 6 sided die looking for a 2, either it will come up or it won’t, just like pulling the Jack of Spades. So, I guess my question is not why am I worng, but why isn’t 50/50 correct?
This is actually a very good point to raise, one that often goes unnoticed in the discussions I’ve seen here about probability. You may or may not be “a math dumbass”, but you’ve at least managed to perceive a problem with a view implicitly, glibly endorsed by many others.
People often think “Whenever there are K possibilities, each has probability 1/K.” This is not necessarily the case. Nothing in the laws of probability demand this. What the laws of probability do demand are that, given an exhaustive list of mutually exclusive possibilities, their probabilities must add up to 1. Therefore, if you already know somehow that each is equally probable, then, each must have probability 1/K. However, if you don’t already know that each is equally probable, then all bets are off, so to speak.
There is no prior reason to suspect that getting a 2 and not getting a 2 on a die are equally probable, nor that getting a Jack of Spades and not getting a Jack of Spades are equally probable. However, there are good symmetry arguments for “getting a 1”, “getting a 2”, “getting a 3”, etc., to all be equally probable, so one can conclude each has probability 1/6. Similarly, there are good symmetry arguments for “getting a Jack of Spades”, “getting a Jack of Clubs”, “getting a 2 of Hearts”, etc., to all be equally probable, so one can conclude each has probability 1/52.
“Either you will or you won’t” doesn’t make the chances 50-50. It’s more likely that you won’t get a 2. Five times more likely, in fact, since there are five ways of not getting a 2, and only one way of getting a 2.
Likewise it’s 51 times more likely that you won’t get the Jack of Spades than that you will.
Do you really think that if you drew a card at random out of the full deck lots of times, it’d be the Jack of Spades half the time? Because that’s what your answer of 50-50 implies.
You are confusing possible outcomes with probabilities. If I said “what is the chance of the sun not rising today” the answer is not 50%, just because it either rises or it doesn’t. Thinking this way would seem to imply 50% probabilities for almost everything, which clearly is not the case.
Ask yourself what 50/50 means. It means if you did something 100 times, you would get one result 50 times and the other result 50 times. Do you really think that if you rolled a normal D6 100 times you’d get 2 50 times and not-2 50 times?
As one more counterexample, suppose that we assume that because you either get a two or you don’t when you roll a six-sided die, that there must be a 50% chance of getting the two. Now what are the odds of rolling a three? By the same reasoning, it must be 50%. What about a one? Again, 50%. But now the probabilities add up to more than 100%–in fact, if you continue this line of reasoning for all possible outcomes, you’ll see that they add up to 300%–and I really can’t make sense of that. So it seems that your initial assumption can’t be right.
50-50 implies 50% probability (i.e, you expect the result half the time). This is obviously not true, because if rolling a “2” had 50% odds, and rolling a “3” had 50% odds, you add the two up and you’re already at 100%. What happened to rolling a 1, 4, 5, & 6?
Probability is trying to put a number as to how likely as possible outcome will happen. Some events will have many possible outcomes.
You roll a die, the possible are outcomes are to get 1, 2, 3, 4, 5 or 6 facing up. You are hoping for a 2 to come up. That is 1 outcome out of six possibilities. So the chance, or likelihood, of getting that number 2 is 1 chance out 6 possible outcomes.
By saying “it will or it won’t” you are misidentifying the possible outcomes. The “will” category has one possible outcome (a 2 facing up) but the “won’t” category actually has five possible outcomes within it (numbers 1, 3, 4, 5, 6 facing up).
last point:
So the “won’t” category actually has more ways to occur, and is therefore more probable to happen (5/6), whereas the “will” category has fewer ways to happen, and is then less probable (1/6).
Say you enter the lottery this weekend. Either you’ll win it or you won’t. Does that make the odds 50:50?
Both the examples you give involve chosing one specific outcome from a number of **equally likely **options.
The die has six faces, all equally likely, and you are looking for one of them. So you will be successful 1 time in 6. That’s 1:6 or 16.66…%
A pack of cards has 52 cards, each equally likely to be drawn. You are looking for just one card. One card gives you the desired result, 51 cards give you the wrong result. The chance is 1 in 52.
Some of you who are simply talking about “the number of ways to occur” or “the number of possible outcomes” are missing the point. The OP has not actually erred in giving a description of the situation as having two possible outcomes. Which is to say, one can always partition outcomes any way one likes; a die has three possible outcomes: 1, prime, or composite. A die has two possible outcomes: 5 or not 5. A die has 6 possible outcomes. A die has 942 possible outcomes, depending on its exact orientation as it lands. Etc.
Colophon is right to bold “equally likely” in his explanation. If you have partitioned outcomes in such a way as that you know each is equally probable, then you can conclude that each of your K outcomes has probability 1/K. But there is nothing intrinsically wrong with partitioning outcomes some other way; indeed, in some cases, one has no choice but to do so. One would just have to use different reasoning after that point to determine the probabilities.
That is to say, the error in the OP is not “I have decided to think of a die as having two possible outcomes”; that is perfectly kosher.
The error in the OP is in making the inference “Where there are K possible outcomes, each has probability 1/K”, without having first established that those outcomes are all equally probable.
One other way to think about it is: Either a two will come up or not implies 50/50. Either an even number will come up or an odd number implies 50/50. But the chances of getting an even number must exceed the chances of getting a two.
These questions are imposible to answer without adding some additional assumption or information. Often when such questions are asked, what they really imply is “What are the chances of getting a 2 on a 6-sided die if all six sides are equally likely to appear?” or “What are the chances of pulling the Jack of Spades from a deck of cards if all cards are equally likely to be chosen?” But those additional assumptions may or may not be true. If you’re dealing with a die that you suspect may be loaded, the probability of rolling a 2 is something you can only figure out by experiment or investigation of that die.
(In fact, there’s some philosophical debate over what probability is in relation to “reality”: Is it just a theoretical, mathematical construct, that we can assign any values to as long as they follow the rules? Does it represent long-run frequencies?)
Toss a hundred coins. Scoop up all the tails . Flip the remainders. Scoop out the tails. Continue until you are down to one. You will find that one coin flipped head 6 or 7 times in a row. Another might have flipped tails 6 times in a row. A fifty /fifty probability should recognize that we get a bell curve. The visualization of 50/50 seems to require a more complete picture.